A continuous random variable with P.D.F f(x) = kx(1 - x), 0 $\mathrm{\le}$ x $\mathrm{\le}$ 1. Find K and determine a number ab such that P(x $\mathrm{\le}$ b) = p(x $\mathrm{\ge}$ b).

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written 9.4 years ago by

teamques10 ★ 70k

modified 3.6 years ago by

starspace973 • 20

For probability density function.

(P.D.F.) $$\int^{\infty }_{-\infty }{f\left(x\right)dx=1}\ \\ \therefore k\ \int^1_0{kx\left(1-x\right)dx=1}\therefore k\int^1_0{\left(x-x^2\right)dx=1}\\ \therefore k{\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}^1_0=1\\\ \therefore k\left\{\left(\frac{1}{2}-\frac{1}{3}\right)-\left(0-0\right)\right\}=1\\ \therefore \frac{k}{6}=1\therefore k=6\\ \therefore f\left(x\right)=6x(1-x)$$

Part II

Given P(x $\mathrm{\le}$ b)=P(x $\mathrm{\ge}$ b)

But P(x $\mathrm{\le}$ b)+P(x $\mathrm{\ge}$ b)=1

$\mathrm{\therefore }$ P(x $\mathrm{\le}$ b)=P(x $\mathrm{\ge}$ b)=$\ \frac{1}{2}$

Consider P(x $\mathrm{\le}$ b)=$\ \frac{1}{2}$ $$\therefore \int^b_0{6x\left(1-x\right)dx=\frac{1}{2}}\\ \therefore \ 6\int^b_0{\left(x-x^2\right)dx=\frac{1}{2}}\\ \therefore 6{\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}^b_0=\frac{1}{2}\\ \therefore \ \left\{\left[\frac{b^2}{2}-\frac{b^3}{3}\right]-\left[0-0\right]\right\}=1\\ \therefore {6b}^2-{4b}^3=1$$

Solving on cal c we get

B=1.366,-0.366,0.5

But 0$\lt$b$<$1 $\mathrm{\therefore }$ b=0.5

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